Problem: A sequence $(a_n)$ is defined as follows: $a_1 = 1,$ $a_2 = \frac{1}{2},$ and
\[a_n = \frac{1 - a_{n - 1}}{2a_{n - 2}}\]for all $n \ge 3.$  Find $a_{120}.$
Explanation: We compute the first few terms:
\[a_1 = 1, \quad a_2 = \frac{1}{2}, \quad a_3 = \frac{1}{4}, \quad a_4 = \frac{3}{4}, \quad a_5 = \frac{1}{2}, \quad a_6 = \frac{1}{3}, \quad a_7 = \frac{2}{3}, \quad a_8 = \frac{1}{2}.\]The sequence appears to be converging to $\frac{1}{2}.$  In fact, every third term appears to be $\frac{1}{2}.$  So we can define a new sequence $(b_n)$ where $b_n = 2a_n - 1.$  Then $a_n = \frac{b_n + 1}{2}.$  Substituting, we get
\[\frac{b_n + 1}{2} = \frac{1 - \frac{1 + b_{n - 1}}{2}}{2 \cdot \frac{1 + b_{n - 2}}{2}}.\]This simplifies to
\[b_n = -\frac{b_{n - 1} + b_{n - 2}}{b_{n - 2} + 1}.\]Note that $b_1 = 1,$ $b_2 = 0,$ and $b_3 = -\frac{1}{2}.$

Suppose $b_n = 0.$  Then
\begin{align*}
b_{n + 1} &= -\frac{b_n + b_{n - 1}}{b_{n - 1} + 1} = -\frac{b_{n - 1}}{b_{n - 1} + 1}, \\
b_{n + 2} &= -\frac{b_{n + 1} + b_n}{b_n + 1} = -b_{n + 1} = \frac{b_{n - 1}}{b_{n - 1} + 1}, \\
b_{n + 3} &= -\frac{b_{n + 2} + b_{n + 1}}{b_{n + 1} + 1} = 0, \\
b_{n + 4} &= -\frac{b_{n + 2}}{b_{n + 2} + 1} = \frac{b_{n + 1}}{1 - b_{n + 1}}.
\end{align*}This tells us if $b_n = 0,$ then $b_{n + 3} = 0.$  Hence, $b_{3m - 1} = 0$ for all $m \ge 1.$

Furthermore, if $b_{n + 1} = -\frac{1}{k},$ then
\[b_{n + 4} = \frac{b_{n + 1}}{1 - b_{n + 1}} = \frac{-1/k}{1 + 1/k} = -\frac{1}{k + 1}.\]Hence, $b_6 = -\frac{1}{3},$ $b_9 = -\frac{1}{4},$ $b_{12} = -\frac{1}{5},$ and so on.  In general,
\[b_{3m} = -\frac{1}{m + 1}.\]Then
\[a_{3m} = \frac{b_{3m} + 1}{2} = \frac{-1/(m + 1) + 1}{2} = \frac{m}{2(m + 1)}.\]In particular,
\[a_{120} = \frac{40}{2(40 + 1)} = \boxed{\frac{20}{41}}.\]